Theorem frgrwopreglem5lem 30840

Description: Lemma for frgrwopreglem5 30841. (Contributed by AV, 5-Feb-2022.)

Hypotheses

Ref	Expression
frgrwopreg.v	⊢ 𝑉 = (Vtx‘𝐺)
frgrwopreg.d	⊢ 𝐷 = (VtxDeg‘𝐺)
frgrwopreg.a	⊢ 𝐴 = {𝑥 ∈ 𝑉 ∣ (𝐷‘𝑥) = 𝐾}
frgrwopreg.b	⊢ 𝐵 = (𝑉 ∖ 𝐴)
frgrwopreg.e	⊢ 𝐸 = (Edg‘𝐺)

Assertion

Ref	Expression
frgrwopreglem5lem	⊢ (((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) ∧ (𝑏 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵)) → ((𝐷‘𝑎) = (𝐷‘𝑥) ∧ (𝐷‘𝑎) ≠ (𝐷‘𝑏) ∧ (𝐷‘𝑥) ≠ (𝐷‘𝑦)))

Distinct variable groups:   𝑥,𝑉   𝑥,𝐴   𝑥,𝐺   𝑥,𝐾   𝑥,𝐷   𝐴,𝑏   𝑥,𝐵   𝑦,𝐷   𝐺,𝑎,𝑏,𝑦,𝑥   𝑦,𝑉

Allowed substitution hints:   𝐴(𝑦, 𝑎)   𝐵(𝑦, 𝑎, 𝑏)   𝐷(𝑎, 𝑏)   𝐸(𝑥, 𝑦, 𝑎, 𝑏)   𝐾(𝑦, 𝑎, 𝑏)   𝑉(𝑎, 𝑏)

Proof of Theorem frgrwopreglem5lem

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frgrwopreg.a	. . . . . 6 ⊢ 𝐴 = {𝑥 ∈ 𝑉 ∣ (𝐷‘𝑥) = 𝐾}
2	1	reqabi 3434	. . . . 5 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝑉 ∧ (𝐷‘𝑥) = 𝐾))
3		fveqeq2 6883	. . . . . . 7 ⊢ (𝑥 = 𝑎 → ((𝐷‘𝑥) = 𝐾 ↔ (𝐷‘𝑎) = 𝐾))
4	3, 1	elrab2 3649	. . . . . 6 ⊢ (𝑎 ∈ 𝐴 ↔ (𝑎 ∈ 𝑉 ∧ (𝐷‘𝑎) = 𝐾))
5		eqtr3 2782	. . . . . . . . 9 ⊢ (((𝐷‘𝑎) = 𝐾 ∧ (𝐷‘𝑥) = 𝐾) → (𝐷‘𝑎) = (𝐷‘𝑥))
6	5	expcom 419	. . . . . . . 8 ⊢ ((𝐷‘𝑥) = 𝐾 → ((𝐷‘𝑎) = 𝐾 → (𝐷‘𝑎) = (𝐷‘𝑥)))
7	6	adantl 487	. . . . . . 7 ⊢ ((𝑥 ∈ 𝑉 ∧ (𝐷‘𝑥) = 𝐾) → ((𝐷‘𝑎) = 𝐾 → (𝐷‘𝑎) = (𝐷‘𝑥)))
8	7	com12 33	. . . . . 6 ⊢ ((𝐷‘𝑎) = 𝐾 → ((𝑥 ∈ 𝑉 ∧ (𝐷‘𝑥) = 𝐾) → (𝐷‘𝑎) = (𝐷‘𝑥)))
9	4, 8	simplbiim 514	. . . . 5 ⊢ (𝑎 ∈ 𝐴 → ((𝑥 ∈ 𝑉 ∧ (𝐷‘𝑥) = 𝐾) → (𝐷‘𝑎) = (𝐷‘𝑥)))
10	2, 9	biimtrid 245	. . . 4 ⊢ (𝑎 ∈ 𝐴 → (𝑥 ∈ 𝐴 → (𝐷‘𝑎) = (𝐷‘𝑥)))
11	10	imp 412	. . 3 ⊢ ((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) → (𝐷‘𝑎) = (𝐷‘𝑥))
12	11	adantr 486	. 2 ⊢ (((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) ∧ (𝑏 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵)) → (𝐷‘𝑎) = (𝐷‘𝑥))
13		frgrwopreg.v	. . . 4 ⊢ 𝑉 = (Vtx‘𝐺)
14		frgrwopreg.d	. . . 4 ⊢ 𝐷 = (VtxDeg‘𝐺)
15		frgrwopreg.b	. . . 4 ⊢ 𝐵 = (𝑉 ∖ 𝐴)
16	13, 14, 1, 15	frgrwopreglem3 30834	. . 3 ⊢ ((𝑎 ∈ 𝐴 ∧ 𝑏 ∈ 𝐵) → (𝐷‘𝑎) ≠ (𝐷‘𝑏))
17	16	ad2ant2r 760	. 2 ⊢ (((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) ∧ (𝑏 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵)) → (𝐷‘𝑎) ≠ (𝐷‘𝑏))
18		fveqeq2 6883	. . . . . 6 ⊢ (𝑥 = 𝑧 → ((𝐷‘𝑥) = 𝐾 ↔ (𝐷‘𝑧) = 𝐾))
19	18	cbvrabv 3422	. . . . 5 ⊢ {𝑥 ∈ 𝑉 ∣ (𝐷‘𝑥) = 𝐾} = {𝑧 ∈ 𝑉 ∣ (𝐷‘𝑧) = 𝐾}
20	1, 19	eqtri 2783	. . . 4 ⊢ 𝐴 = {𝑧 ∈ 𝑉 ∣ (𝐷‘𝑧) = 𝐾}
21	13, 14, 20, 15	frgrwopreglem3 30834	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝐷‘𝑥) ≠ (𝐷‘𝑦))
22	21	ad2ant2l 759	. 2 ⊢ (((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) ∧ (𝑏 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵)) → (𝐷‘𝑥) ≠ (𝐷‘𝑦))
23	12, 17, 22	3jca 1146	1 ⊢ (((𝑎 ∈ 𝐴 ∧ 𝑥 ∈ 𝐴) ∧ (𝑏 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵)) → ((𝐷‘𝑎) = (𝐷‘𝑥) ∧ (𝐷‘𝑎) ≠ (𝐷‘𝑏) ∧ (𝐷‘𝑥) ≠ (𝐷‘𝑦)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2145   ≠ wne 2955  {crab 3412   ∖ cdif 3896  ‘cfv 6528  Vtxcvtx 29493  Edgcedg 29544  VtxDegcvtxdg 29965

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-iota 6484  df-fv 6536

This theorem is used by:  frgrwopreglem5  30841